Enderton. Axiom of choice; fix theorem 3J.
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\chapter{Reference}%
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\label{chap:reference}
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\section{\partial{Axiom of Choice, First Form}}%
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\label{ref:axiom-of-choice-1}
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For any relation $R$ there is a function $H \subseteq R$ with
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$\dom{H} = \dom{R}$.
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\section{\defined{Composition}}%
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\label{ref:composition}
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@ -3333,7 +3339,7 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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one-to-one.
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\item There exists a function $H \colon B \rightarrow A$ (a "right inverse")
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such that $F \circ H$ is the identity function $I_B$ on $B$ iff $F$ maps
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$A$ onto $B$.
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$A$ \textit{onto} $B$.
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\end{enumerate}
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\end{theorem}
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@ -3349,55 +3355,117 @@ For any one-to-one function $F$, $F^{-1}$ is also one-to-one.
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\subparagraph{($\Rightarrow$)}%
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Suppose $G \colon B \rightarrow A$ is a function satisfying
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$G \circ F = I_A$.
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To prove function $F$ is one-to-one, it is sufficient to prove $F$ is
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single-rooted.
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Let $G \colon B \rightarrow A$ such that $G \circ F = I_A$.
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All that remains is to prove $F$ is single-valued.
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Let $y \in \ran{F}$.
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Then there there exists some $x_1 \in \dom{F}$ such that $F(x_1) = y$.
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Suppose there exists another $x_2 \in \dom{F}$ such that $F(x_2) = y$.
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Then $G(F(x_1)) = G(y) = G(F(x_2))$.
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But $G \circ F = I_A$ so $x_1 = x_2$.
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Thus, for all $y \in \ran{F}$, there exists only one $x$ such that
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$\left< x, y \right> \in F$.
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That is, $F$ is single-rooted as expected.
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By definition of the \nameref{ref:range} of a function, there exists some
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$x$ such that $\left< x, y \right> \in F$.
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Suppose $x_1, x_2 \in \dom{F}$ such that
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$\left< x_1, y \right>, \left< x_2, y \right> \in F$.
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Then $F(x_1) = F(x_2) = y$.
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Then $G(F(x_1)) = G(F(x_2))$ implies $I_A(x_1) = I_A(x_2)$.
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Thus $x_1 = x_2$.
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\subparagraph{($\Leftarrow$)}%
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Suppose $F$ is one-to-one.
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Then by \nameref{sub:lemma-1}, $F^{-1} \colon B \rightarrow A$ is also
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one-to-one.
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We show that $F^{-1} \circ F = I_A$.
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Let $x \in \dom{F}$.
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By \nameref{sub:theorem-3h}, $(F^{-1} \circ F)(x) = F^{-1}(F(x))$.
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By \nameref{sub:theorem-3g}, $F^{-1}(F(x)) = x$.
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Thus $F^{-1} \circ F = I_A$.
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Let $F$ be one-to-one.
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Since $A$ is nonempty, there exists some $a \in A$.
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Let $G \colon B \rightarrow A$ be given by
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$$G(y) = \begin{cases}
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F^{-1}(y) & \text{if } y \in \ran{F} \\
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a & \text{otherwise}.
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\end{cases}$$
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$G$ is a function by virtue of \nameref{sub:lemma-1} and choice of mapping
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for all values $y \not\in \ran{F}$.
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Furthermore, for all $x \in A$, $F(x) \in \ran{F}$.
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Thus $(G \circ F)(x) = G(F(x)) = F^{-1}(F(x)) = x$ by
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\nameref{sub:theorem-3g}.
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\paragraph{(b)}%
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We prove there exists a function $H \colon B \rightarrow A$ such that
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$F \circ H = I_B$ if and only if $F$ maps $A$ onto $B$.
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$F \circ H = I_A$ if and only if $F$ maps $A$ onto $B$.
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\subparagraph{($\Rightarrow$)}%
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Suppose $H \colon B \rightarrow A$ is a function satisfying
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$F \circ H = I_B$.
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To prove $F$ maps $A$ onto $B$, we must prove $\ran{F} = B$.
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But for all $y \in B$, $F(H(y)) = y$.
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Thus $y \in \ran{F}$ meaning $B \subseteq \ran{F}$.
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Since $F$ maps $A$ into $B$ by hypothesis, $\ran{F} \subseteq B$.
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Thus $\ran{F} = B$ as expected.
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Suppose $H \colon B \rightarrow A$ such that $F \circ H = I_A$.
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All that remains is to prove $\ran{F} = B$.
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Note that $\ran{F} \subseteq B$ by hypothesis.
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Let $y \in B$.
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But $F(H(y)) = y$ meaning $y \in \ran{F}$.
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Thus $B \subseteq \ran{F}$.
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Since $\ran{F} \subseteq B$ and $B \subseteq \ran{F}$, $\ran{F} = B$.
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\subparagraph{($\Leftarrow$)}%
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Suppose $F$ maps $A$ onto $B$.
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That is, $\ran{F} = B$.
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We show that $F^{-1} \colon B \rightarrow A$ satisfies
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$F \circ F^{-1} = I_B$.
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Let $y \in \ran{F}$.
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By \nameref{sub:theorem-3h}, $(F \circ F^{-1})(y) = F(F^{-1}(y))$.
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By \nameref{sub:theorem-3g}, $F(F^{-1}(y)) = y$.
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Thus $F \circ F^{-1} = I_B$.
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Suppose $F$ maps $A$ \textit{onto} $B$.
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By definition of maps onto, $\ran{F} = B$.
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Then for all $y \in B$, there exists some $x \in A$ such that
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$\left< x, y \right> \in F$.
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Notice though that $F^{-1}[\{y\}]$ may not be a singleton set.
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Then there is no obvious way to \textit{choose} an element from each
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preimage to form a function.
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By the \nameref{ref:axiom-of-choice-1}, there exists a function
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$H \subseteq F^{-1}$ such that $\dom{H} = \dom{F^{-1}} = B$.
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For all $y \in B$, $\left< y, H(y) \right> \in H \subseteq F^{-1}$
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meaning $\left< H(y), y \right> \in F$.
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Thus $F(H(y)) = y$ as expected.
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\end{proof}
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\subsection{\unverified{Theorem 3K}}%
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\label{sub:theorem-3k}
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\begin{theorem}[3K]
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The following hold for any sets. ($F$ need not be a function.)
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\begin{enumerate}[(a)]
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\item The image of a union is the union of the images:
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$$F[A \cup B] = F[A] \cup F[B] \quad\text{and}\quad
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F\left[\bigcup{\mathscr{A}}\right] =
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\bigcup\;\{F[A] \mid A \in \mathscr{A}\}.$$
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\item The image of an intersection is included in the intersection of the
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images:
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$$F[A \cap B] \subseteq F[A] \cap F[B] \quad\text{and}\quad
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F\left[\bigcap\mathscr{A}\right] \subseteq
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\bigcap\;\{F[A] \mid A \in \mathscr{A}\}$$
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for nonempty $\mathscr{A}$.
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Equality holds if $F$ is single-rooted.
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\item The image of a difference includes the difference of the images:
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$$F[A] - F[B] \subseteq F[A - B].$$
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Equality holds if $F$ is single-rooted.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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TODO
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\end{proof}
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\subsection{\unverified{Corollary 3L}}%
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\label{sub:corollary-3l}
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\begin{theorem}[3L]
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For any function $G$ and sets $A$, $B$, and $\mathscr{A}$:
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\begin{align*}
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G^{-1}\left[\bigcup\mathscr{A}\right]
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& = \bigcup\;\{G^{-1}[A] \mid A \in \mathscr{A}\}, \\
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G^{-1}\left[\bigcap\mathscr{A}\right]
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& = \bigcap\;\{G^{-1}[A] \mid A \in \mathscr{A}\}
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\text{ for } \mathscr{A} \neq \emptyset, \\
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G^{-1}[A - B]
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& = G^{-1}[A] - G^{-1}[B].
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\end{align*}
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\end{theorem}
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\begin{proof}
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TODO
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\end{proof}
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